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Heat Sinks - 1Copyright © by John Wiley & Sons 2003
Heat Sinks and Component Temperature Control
Lecture Notes
Heat Sinks - 2Copyright © by John Wiley & Sons 2003
• All components, capacitors, inductors and transformers, and semiconductor devicesand circuits have maximum operating temperatures specified by manufacturer.
• Component reliability decreases with increasing temperature.Semiconductorfailure rate doubles for every 10 - 15 °C increase in temperature above 50 °C(approx. rule-of-thumb).
• High component operating temperatures have undesirable effects on components.
Need for Component Temperature Control
Capacitors
Electrolyte evaporationrate increasessignificantly with temperature increasesand thus shortenslifetime.
Magnetic Components
• Losses (at constant power input) increase above 100 °C
• Winding insulation (lacqueror varnish) degrades above100 °C
Semconductors
• Unequal power sharing in paralleled or seriesed devices.
• Reduction in breakdown voltage in some devices.
• Increase in leakage currents.
• Increase in switching times.
Heat Sinks - 3Copyright © by John Wiley & Sons 2003
• Control voltages across and current through components via good design practices.
• Snubbers may be required for semiconductor devices.
• Free-wheeling diodes may be needed with magnetic components.
• Use components designed by manufacturers to maximize heat transfer via convection and radiation from component to ambient.
• Short heat flow paths from interior to component surface and large component surface area.
• Component user has responsibility to properly mount temperature-critical components on heat sinks.
• Apply recommended torque on mounting bolts and nuts and use thermal greasebetween component and heat sink.
• Properly design system layout and enclosure for adequate air flow so that heat sinks can operate properly to dissipate heat to the ambient.
Temperature Control Methods
Heat Sinks - 4Copyright © by John Wiley & Sons 2003
Heat Conduction Thermal Resistance
• Generic geometryof heat flow viaconduction
Pcond
Temperature = T1Temperature = T2
d
h
b
T > T12
heat flow direction
• Heat flow Pcond [W/m2] = l A (T2 - T1) / d = (T2 - T1) / Rqcond
• Thermal resistance Rqcond = d / [l A]
• Cross-sectional area A = hb
• l = Thermal conductivity has units of W-m-1-°C-1 (lAl = 220 W-m-1-°C-1 ).• Units of thermal resistance are °C/W
Heat Sinks - 5Copyright © by John Wiley & Sons 2003
Thermal Equivalent Circuits
• Heat flow througha structure composedof layers of differentmaterials.
• Thermal equivalent circuit simplifies calculation of temperatures in various partsof structure.
PR qsaqcsR
qjcR
Junction Case Sink Ambient
jT cT sT aT++++
----
Chip Tj
Case Tc
Isolation pad
Heat sink Ts
Ambient Temperature Ta
• Ti = Pd (Rqjc + Rqcs + Rqsa) + Ta
• If there parallel heat flow paths,then thermal resistances of theparallel paths combine as doelectrical resistors in parallel.
Heat Sinks - 6Copyright © by John Wiley & Sons 2003
Transient Thermal Impedance
• Heat capacity per unit volume Cv = dQ/dT [Joules /°C] prevents short duration highpower dissipation surges from raising component temperature beyond operating limits.
• Transient thermal equivalentcircuit. Cs = CvV where V is thevolume of the component.
P(t)qR
jT (t)
aT
Cs
P(t)
t
qRPo
qt tSlope = 0.5
log Z (t)q
• Transient thermal impedance Zq(t) = [Tj(t) - Ta]/P(t)
• tq = π Rq Cs /4 = thermal timeconstant
• Tj(t = tq) = 0.833 Po Rq
Heat Sinks - 7Copyright © by John Wiley & Sons 2003
Application of Transient Thermal Impedance
• Symbolic response for a rectangular power dissipation pulse P(t) = Po {u(t) - u(t - t1)}.
P(t)
t
qRPo
qt tt1 t1
Z (t)q
1q-Z (t - t )
net response
q-R
• Tj(t) = Po { Zq(t) - Zq(t - t1) }
• Symbolic solution for half sine powerdissipation pulse.
• P(t) = Po {u(t - T/8) - u(t - 3T/8)} ; areaunder two curves identical.
• Tj(t) = Po { Zq(t - T/8) - Zq(t - 3T/8) }T/8 3T/8
T/2
Po
t
P(t)
Equivalent rectangular pulse
Half sine pulse
Heat Sinks - 8Copyright © by John Wiley & Sons 2003
Zq for Multilayer StructuresP(t)
SiliconCopper mount
Heat sink
TjCuT
cTaT
• Multilayer geometry
• Transient thermalequivalent circuit
Tj CuT cT
aT
C (Si)s C (Cu)s C (sink)sP(t)
R (sink)qR (Cu)
qR (Si)
q
R (Si)q
R (Cu)q
R (Si)q
+
R (sink)q
R (Cu)q
R (Si)q
+ +
log(t)
log[Z (t)]q
t (Si)q
t (sink)qt (Cu)
q
• Transient thermalimpedance (asymptotic)of multilayer structureassuming widelyseparated thermal timeconstants.
Heat Sinks - 9Copyright © by John Wiley & Sons 2003
Heat Sinks• Aluminum heat sinks of various shapes and sizes widely available for cooling components.
• Often anodized with black oxide coating to reduce thermal resistance by up to 25%.
• Sinks cooled by natural convection have thermal time constants of 4 - 15 minutes.
• Forced-air cooled sinks have substantially smaller thermal time constants, typicallyless than one minute.
• Choice of heat sink depends on required thermal resistance, Rqsa, which is determined byseveral factors.
• Maximum power, Pdiss, dissipated in the component mounted on the heat sink.
• Component's maximum internal temperature, Tj,max • Component's junction-to-case thermal resistance, Rqjc.
• Maximum ambient temperature, Ta,max.
• Rqsa = {Tj,max - Ta,max}Pdiss - Rqjc
• Pdiss and Ta,max determined by particular application.
• Tj,max and Rqjc set by component manufacturer.
Heat Sinks - 10Copyright © by John Wiley & Sons 2003
Radiative Thermal Resistance
• Stefan-Boltzmann law describes radiative heat transfer.
• Prad = 5.7x10-8 EA [( Ts)4 -( Ta)4 ] ; [Prad] = [watts]
• E = emissivity; black anodized aluminum E = 0.9 ; polished aluminum E = 0.05
• A = surface area [m2]through which heat radiation emerges.
• Ts = surface temperature [°K] of component. Ta = ambient temperature [°K].
• (Ts - Ta )/Prad = R q,rad = [Ts - Ta][5.7EA {( Ts/100)4 -( Ta/100)4 }]-1
• Example - black anodized cube of aluminum 10 cm on a side. Ts = 120 °C and
Ta = 20 °C
• Rq,rad = [393 - 293][(5.7) (0.9)(6x10-2){(393/100)4 - (293/100)4 }]-1
• Rq,rad = 2.2 °C/W
Heat Sinks - 11Copyright © by John Wiley & Sons 2003
Convective Thermal Resistance
• Pconv = convective heat loss to surrounding air from a vertical surface at sea level havinga height dvert [in meters] less than one meter.
• Pconv = 1.34 A [Ts - Ta]1.25 dvert-0.25
• A = total surface area in [m2]
• Ts = surface temperature [°K] of component. Ta = ambient temperature [°K].
• [Ts - Ta ]/Pconv = Rq,conv = [Ts - Ta ] [dvert]0.25[1.34 A (Ts - Ta )1.25]-1
• Rq,conv = [dvert]0.25 {1.34 A [Ts - Ta]0.25}-1
• Example - black anodized cube of aluminum 10 cm on a side. Ts = 120 °C and Ta = 20 °C.
• Rq,conv = [10-1]0.25([1.34] [6x10-2] [120 - 20]0.25)-1
• Rq,conv = 2.2 °C/W
Heat Sinks - 12Copyright © by John Wiley & Sons 2003
Combined Effects of Convection and Radiation
• Heat loss via convection and radiation occur in parallel.
• Steady-state thermal equivalent circuit
• Rq,sink = Rq,rad Rq,conv / [Rq,rad + Rq,conv]
• Example - black anodized aluminum cube 10 cm per side
• Rq,rad = 2.2 °C/W and Rq,conv = 2.2 °C/W
• Rq,sink = (2.2) (2.2) /(2.2 + 2.2) = 1.1 °C/W
P R q,convq,radR
sT
aT